Question

The loop in the figure below is being pushed into the 0.20 T magnetic field at 50 m/s. The resistance of the loop is 0.10 , and its width d = 5.4 cm. What are the direction and the magnitude of the current in the loop? (in units of A)

Answer 1

Concepts and reason

The concept used to this problem is induced EMF in the loop. Initially, from the expression of the EMF, calculate the value of the EMF. Later, apply Ohm's law to determine the current in the loop. Finally, determine the direction of the current in the loop.

Fundamental

The expression for the EMF is as follows:

\(\varepsilon=B l v\)

Here, \(\mathrm{B}\) is the magnetic field, \(l\) is the length, and \(\mathrm{v}\) is the velocity vector. The expression for Ohm's law is as follows:

\(\varepsilon=I R\)

Here, \(\varepsilon\) is the potential difference, \(I\) is current, and \(\mathrm{R}\) is the resistance.

The expression of the induced emf in the loop is as follows:

\(\varepsilon=B l v\)

Substitute \(0.20 \mathrm{~T}\) for \(\mathrm{B}, 0.054 \mathrm{~m}\) for \(\mathrm{l}\), and \(50 \mathrm{~m} / \mathrm{s}\) for \(v\) in the equation \(\varepsilon=B l v .\)

\(\varepsilon=(0.20 \mathrm{~T})(0.054 \mathrm{~m})(50 \mathrm{~m} / \mathrm{s})\)

\(=0.54 \mathrm{~V}\)

Explanation | Hint for next step

The magnitude of the induced emf in the loop is calculated by the expression of the induced emf in the loop.

Rearrange the equation of Ohm's law for I. \(I=\frac{\varepsilon}{R}\)

Substitute \(0.54 \mathrm{~V}\) for \(\varepsilon\) and \(0.10 \Omega\) for \(\mathrm{R}\) in the equation \(I=\frac{\varepsilon}{R}\).

\(I=\frac{0.54 \mathrm{~V}}{0.10 \Omega}\)

\(=5.40 \mathrm{~A}\)

Explanation | Hint for next step

According to ohm's law, the emf induced across the loop is, \(\varepsilon=I R\)

Rearrange the above equation for \(I\). \(I=\frac{\varepsilon}{R}\)

As the loop moves more into the field, the flux is increasing. Therefore, the induced field opposes the applied field, and the applied field points out. Thus, the induced field points inside. Hence, the direction of current in the loop is Clockwise.

The direction of the velocity is upward, and the magnetic field is on the page. Thus, the force should be perpendicular to both velocity and field.

The magnitude of the current in the loop is equal to 5.40 A and the direction of the current is in the clockwise direction.